Representing real numbers on the number line

(-\frac{3}{2}=-1.5), so it lies between (-2) and (-1). In exams, the decimal form of a negative fraction helps.

From \(x^2-25=0\), we get \(x^2=25\). Therefore, \(x=5\) or \(x=-5\), so the solutions lie at −5 and 5 on the number line. Option B incorrectly includes 25; 25 is not a value of \(x\) satisfying the equation. Exam tip: When taking the square root of a positive number, consider both the positive and negative values.

From (2x-1=0), (x=\frac{1}{2}), and from (x+4=0), (x=-4). In exams, solve each linear factor separately.

(\sqrt{2}) is about (1.414), so the order is (0,\frac{1}{2},\sqrt{2},2). In exams, increasing order is from left to right.

(-\sqrt{2}) is about (-1.414) and (-\frac{3}{2}=-1.5), so (x) is greater. In exams, for negative numbers, the point with smaller distance from (0) may lie to the right.

Every real number corresponds to exactly one point on the number line. Irrational numbers such as \(\sqrt{2}\) have non-terminating, non-repeating decimals, so B and D are false. Exam tip: check whether an endless decimal repeats before classifying it.

Since \(5^2=25<26<36=6^2\), taking positive square roots gives \(5<\sqrt{26}<6\). Therefore, \(\sqrt{26}\) lies between 5 and 6. Exam tip: Compare the number with the two nearest consecutive perfect squares to locate its square root.

Each part is (\frac{3}{12}=\frac{1}{4}), so the seventh point is (\frac{7}{4}). Divide the total length by equal parts.

Since (-\sqrt{16}=-4), the distance between the two points is the absolute value of their difference: (|2.5-(-4)|=|6.5|=6.5). Therefore, 6.5 is correct. The value 5.5 results from an incorrect subtraction of (-4) and (2.5). Exam tip: Distance on a number line is always the absolute value of the difference, so it cannot be negative.

(\sqrt{6}) is about (2.45), so (-\sqrt{6}) is about (-2.45). (-2.4) is greater, so it is to the right.

The midpoint of two numbers is their average: \(\frac{-1.6+2.8}{2}=\frac{1.2}{2}=0.6\). Therefore, option B is correct. Option C is only the sum of the two numbers, not their midpoint. In the exam, remember to divide the sum by 2 to find the midpoint.

The governing idea is that equal numerical values represent the same point on a number line. Convert the terminating decimal exactly: 2.125=2125/1000. Dividing numerator and denominator by 125 gives 17/8. Alternatively, 2.125=2+0.125=2+1/8=16/8+1/8=17/8. Therefore option C is correct. The distractors have different values: 15/8=1.875, 16/8=2, and 19/8=2.375. Thus only 17/8 is exactly equivalent to 2.125; estimation is unnecessary because the decimal can be converted without rounding.

(\sqrt{45}) is about (6.7), so it is closer to (7). Nearby perfect squares (36) and (49) help in estimation.

Moving left on the number line means subtracting the distance from the starting coordinate. Thus, the final coordinate is 1.25 - 2.75 = -1.5. Option C gives only the magnitude of the difference and ignores the negative direction. Exam tip: add for movement to the right and subtract for movement to the left.

(\frac{4}{3}\approx1.33) and (\sqrt{2}\approx1.41), so the order is (\frac{4}{3}<\sqrt{2}<1.5). Estimate values for comparison.

Since (7^2<50<8^2), (\sqrt{50}) lies between (7) and (8). Use perfect squares to decide the interval.

Since \\(4<5<9\\), we have \\(2<\\sqrt{5}<3\\). Adding 2 to all parts gives \\(4<2+\\sqrt{5}<5\\). Therefore, \\(2+\\sqrt{5}\\) lies between 4 and 5. Exam tip: Compare the number under the square root with consecutive perfect squares instead of finding a decimal approximation first.

The total distance is (2), so each part is (\frac{2}{8}=\frac{1}{4}). Find the distance and divide by equal parts.

After the third step, the value is (-1+3\times0.25=-0.25). For equal steps, multiply step number by step length.

The governing concept is comparison of real numbers on a number line. First locate the irrational endpoint: 1.7²=2.89 and 1.8²=3.24, so √3 is approximately 1.732. Hence the open interval between √3 and 2 is approximately (1.732,2). The value 1.8 lies inside this interval, so option B is correct. Number 1.6 is less than √3, while 2.1 and 3 are greater than 2. Therefore they cannot lie strictly between the endpoints. Comparing squares gives a dependable bound and avoids incorrectly replacing √3 by a rough integer estimate.